A planet does not simply orbit its star. Both orbit their common center of mass, so the star makes a small, slow orbit of its own. That motion shows up as a tiny Doppler shift in the star's spectrum, the same method as for spectroscopic binaries with one invisible partner. In 1995 Michel Mayor and Didier Queloz found 51 Pegasi b this way, the first planet around a Sun-like star.
💡 Why it works
Momentum about the center of mass balances: $M_\star v_\star = m_pv_p$. For a light planet on a circular orbit, Kepler's third law gives the planet's speed: $a^3 = GM_\star P^2/(4\pi^2)$ and $v_p = 2\pi a/P = (2\pi GM_\star/P)^{1/3}$. We see only the part of the star's velocity along our line of sight, $K = v_\star\sin i$, where $i$ is the tilt of the orbit ($90^\circ$ is edge-on). Putting these together, $$K = \frac{m_p\sin i}{M_\star}\left(\frac{2\pi GM_\star}{P}\right)^{1/3}, \qquad m_p\sin i = K\,M_\star^{2/3}\left(\frac{P}{2\pi G}\right)^{1/3}.$$
🔑 Key idea
The radial-velocity method measures $m_p\sin i$, the planet's minimum mass. The true mass is $m_p = (m_p\sin i)/\sin i$, which is larger unless the orbit is edge-on.
A heavy star and a light companion. The star's small circle and small velocity swing are what the radial-velocity method detects.
Worked examples
✏️ Worked example: The Sun's wobble from Jupiter
Jupiter ($1.90 \times 10^{27}$ kg) orbits the Sun ($1.99 \times 10^{30}$ kg) at 13.1 km/s. How fast does the Sun move around their center of mass?
$v_\odot = (m_J/M_\odot)\,v_J$.
The mass ratio is $1.90 \times 10^{27}/1.99 \times 10^{30} = 9.55 \times 10^{-4}$, so $v_\odot = 9.55 \times 10^{-4} \times 13{,}100$ m/s.
Answer: About 12.5 m/s, about as fast as the fastest human sprinters, repeating every 11.9 years. Earth moves the Sun by only 9 cm/s.
✏️ Worked example: 51 Pegasi b
51 Pegasi ($1.11\,M_\odot$) wobbles with $K = 56$ m/s and period 4.23 days. Find $m\sin i$ in Jupiter masses.
$P = 3.65 \times 10^{5}$ s, so $(P/2\pi G)^{1/3} = 9.55 \times 10^{4}$ in SI units.
$m\sin i = 56 \times 1.70 \times 10^{20} \times 9.55 \times 10^{4} = 9.1 \times 10^{26}$ kg.
Answer: About $0.48\,M_\text{J}$, a Jupiter-like planet orbiting closer to its star than Mercury does to the Sun.
✏️ Worked example: Adding the tilt
A planet has $m\sin i = 2.0\,M_\text{J}$. Measurements of the star's position on the sky show the orbit is tilted $i = 30^\circ$ from face-on. What is its true mass?
$m = (m\sin i)/\sin i = 2.0/\sin 30^\circ$.
$= 2.0/0.50$.
Answer: $4.0\,M_\text{J}$, twice the minimum.
⚠️ Common mistake
Treating $m\sin i$ as the planet's mass. Doppler shifts see only motion toward or away from us. A planet on a nearly face-on orbit barely shifts its star's lines, so it can look light when it is heavy. The radial-velocity mass is a lower limit. A planet that also transits has $i \approx 90^\circ$, so for it $m\sin i$ is close to the true mass. Check: $\sin i \le 1$, so the true mass can never be less than $m\sin i$.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.